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\(a,x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x-\frac{61}{8}=\frac{5}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)
\(b,x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x+\frac{43}{5}=\frac{37}{4}\)
=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)
\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)
=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)
=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)
d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)
=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)
=> \(\frac{98x}{198}=99\)
=> 98x = 99 . 198
=> 98x = 19602
=> x = 19602 : 98 = 9801/49
a) \(x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{71}{8}\)
b) \(x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x=\frac{37}{4}-\frac{61}{8}\)
=> \(x=\frac{13}{8}\)
c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)
=> \(x-\frac{61}{8}=3.\frac{1}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}\)
=> \(x=\frac{73}{8}\)
d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)
=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)
=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)
=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)
=> \(x.\frac{98}{99}=198\)
=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)
Ta co : 100-99+98-97+...+4-3+2-1
=(100-99)+(98-97)+...+(4-3)+(2-1)
=1+1+...+1+1
Ta lai co: Tu 1->100 co: (100-1)÷1+1=100 so
Co so cap la: 100÷2=50 cap
=1×50
=50
Dãy số trên có SSH là:100 số
Ta ghép 2 số vào 1 nhóm đc 100:2=50(nhóm)
Ta có:(100-99)+(98-97)+...+(4-3)+(2-1)
1+1+1+1+...+1
50.1=50
\(A=3^1+3^4+3^7+...+3^{100}\)
\(A=\left(3^1+3^4+3^7+3^{10}\right)+...+\left(3^{91}+3^{94}+3^{97}+3^{100}\right)\)
\(A=\left(3^1+3^4+3^7+3^{10}\right)+...+3^{96}.\left(3^1+3^4+3^7+3^{10}\right)\)
\(A=\left(3^1+3^4+3^7+3^{10}\right).\left(1+...+3^{96}\right)\)
\(A=61320.\left(1+...+3^{96}\right)\)
\(A=7665.8.\left(1+...+3^{96}\right)⋮8\)
\(\Rightarrow A=3^1+3^4+3^7+...+3^{100}⋮8\)
a, 5x - 1 = 13
=> 5x = 14
=> x = 14/5
b,(x - 2) = 0
=> x - 2 = 0
=> x = 2
c, 5(x - 7) + 8 = 0
=>5(x - 7) = -8
=>x -7 = -8/5 = -1,6
=>x = 5,4
d, (x - 19).4 = 36
=>x - 19 = 9
=>x = 28
e, 3(x - 7) - 2 = 4
=> 3(x - 7) = 6
=> x - 7 = 2
=> x = 9
a)15/8+3/4-5/12
=45+18-10/24
=53/24
b)11/24.12/33+5/6
=11.12/12.2.11.3+5/6
=1/6+5/6
=6/6=1
c)15/8+7/24:5/8
=15/8+7/24.8/5
=15/8+7.8/3.8.5
=15/8+7/15
=đề sai, nếu đúng thì như này
=8/15+7/15
=15/15=1
a: =-3/4+1/2-1/13+3/13=-1/4+2/13=-13/52+8/52=-5/52
b: =10/11+1/11-1/8=1-1/8=7/8
c: =4(2,86+3,14)-30,05+9x0,75
=24-30,05+6,75
=0,7
a)\(4^{10}.8^{15}.\left(2^{10}\right)^2.\left(2^{15}\right)^3=2^{20}.2^{30}=2^{50}\)
b)\(8^2.25^3=\left(2^2\right)^3.\left(5^3\right)^2=2^6.5^6=\left(2.5\right)^6=10^6\)
de mh giai thik : cau a ) vi chung chua cung co so nen ta doi ra thanh co so giong . nhau ta chi viec lay so mu cong voi nhau
cau b ) ta không thể đổi ra cùng cơ số được nên ta đổi ra cung so mu. thi ta chi viec nhan hai co so voi nhau giu nguyen mu
sai chỗ nào thì mấy bạn sữa giùm mình nha mình mới học lớp 5 ak
nhanh len may ban oi\